How do I make a phase diagram for water?

By knowing where the normal boiling and freezing points are (at #"1 atm"#), critical point and triple point are, and the slope of the liquid-solid, liquid-vapor, and solid-vapor coexistence curves.

Note that the phase diagram is simply a pressure vs. temperature graph.

We know that #T_f = 0^@ "C"# at #"1 atm"# and #T_b = 100^@ "C"# at #"1 atm"# are the normal freezing and boiling points, respectively.

The critical point is when the liquid and vapor exist at the same time, at some #(P_c,T_c)# on a #P# vs. #T# graph. For water, #P_c = "218.3 atm"# and #T_c = 374.2^@ "C"#.

The triple point is when the solid, liquid, and vapor exist at the same time, at some #(P_"trpl", T_"trpl")#. For water, #T_"trpl" = 0.01^@ "C"#, and #P_"trpl" = "0.006032 atm"#.

The Clapeyron equation describes the slope of a coexistence curve, #(dP)/(dT) = (DeltabarH_"trs")/(TDeltabarV_("trs"))#. We omit the derivation, but:

• #DeltabarH_"trs"# is the molar enthalpy of the phase transition.
• #DeltabarV_"trs"# is the change in molar volume due to the phase transition.
• #T# is the temperature in #"K"#.

Using the Clapeyron equation:

• For the liquid-solid coexistence curve in the phase diagram of water, #(dP)/(dT) < 0#, since #DeltabarV_((l)->(s)) > 0#, and #DeltabarH_"frz" < 0#, which indicates that water expands when it freezes (this is unusual).
• For the liquid-vapor coexistence curve, #(dP)/(dT) > 0#, since #DeltabarV_((l)->(g)) > 0#, and #DeltabarH_"vap" > 0#, which indicates that water expands when it vaporizes (as usual).
• For the solid-vapor coexistence curve, #(dP)/(dT) > 0#, since #DeltabarV_((s)->(g)) > 0#, and #DeltabarH_"sub" > 0#, which indicates that water expands when it sublimates (clearly).

Put all that together into a phase diagram:

Simply note where each point is, and you should be able to at least sketch its general shape. Note that the phase diagram is simply a pressure vs. temperature graph.